A)Exons 2 and 3 must have the same number of nucleotides. B)Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer). C)Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2). D)Exons 2 and 3 must have different numbers of nucleotides. Multiple Choice

Statement C is the only one that is necessarily true for exons 2 and 3. It is also true for exons 7 and 8. While statements A and B could be true, they don’thave to be. Because the protein sequence is the same in segments of the mRNA that correspond to exons 1 and 10, neither choice of alternative exons (2 versus 3, or 7 versus 8) can alter the reading frame. To maintain the normal reading frame—whatever that is—the alternative exons must have a number of nucleotides that when divided by 3 (the number of nucleotides in a codon) give the same remainder. Since the sequence of the a-tropomyosin gene is known, it is possible to check to see the actual state of affairs. Exons 2 and 3 both contain the same number of nucleotides, 126, which is divisible by 3 with no remainder.

human didn't come from monkeys man i hate these stupid teachers